Question

$\boxed{\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \: dx$

Let assume that,

$\rm :\longmapsto\:I = \displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \: dx - - - (1)$

We know, Variable changing property of integrals

$\red{\boxed{ \tt{ \: \displaystyle \int\limits_{a}^{b} \: f(x) \: dx \: = \: \displaystyle \int\limits_{a}^{b} \: f(y) \: dy \: }}}$

So, above integral can also be rewritten as

$\rm :\longmapsto\:I = \displaystyle \int\limits_{- \infty}^{\infty} e^{-y^2} \: dy - - - (2)$

Now, multiply equation (1) and (2), we get

$\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} dx \times \displaystyle \int\limits_{- \infty}^{\infty} e^{-y^2} \: dy$

can be rewritten as

$\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty}\bigg(\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \times e^{-y^2} \:dx\bigg) \: dy$

$\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty}\bigg(\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2 - {y}^{2} } \:dx\bigg) \: dy$

$\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty}\bigg(\displaystyle \int\limits_{- \infty}^{\infty} e^{-(x^2 + {y}^{2})} \:dx\bigg) \: dy$

Now, we can use polar coordinates to evaluate this integral.

Let we substitute

$\red{\rm :\longmapsto\: {x}^{2} + {y}^{2} = {r}^{2} }$

where,

$\red{\rm :\longmapsto\:x = r \: cos\theta }$

and

$\red{\rm :\longmapsto\:y = r \: sin\theta }$

It represents a circle whose center is (0, 0) and radius is r units.

So,

$\red{\boxed{ \tt{ \: Limits \:of \: x \: changes \: to \: \theta \: varies \: from \: 0 \: to \: 2\pi \: }}}$

and

$\red{\rm :\longmapsto\:dx \: dy \: = \: rd \theta \: dr}$

So, above integral can be rewritten as

$\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{0}^{\infty}\bigg(\displaystyle \int\limits_{0}^{2\pi} d\theta \bigg) \: e^{- {r}^{2} } \:rdr$

can be rewritten as

$\rm :\longmapsto\: {I}^{2} = \displaystyle \int\limits_{0}^{\infty}\bigg(\theta \bigg)_{0}^{2\pi} \: e^{- {r}^{2} } \:rdr$

$\rm :\longmapsto\: {I}^{2} = \displaystyle \int\limits_{0}^{\infty}\bigg(2\pi - 0 \bigg) \: e^{- {r}^{2} } \:2rdr$

$\rm :\longmapsto\: {I}^{2} = \displaystyle \int\limits_{0}^{\infty}\bigg(2\pi \bigg) \: e^{- {r}^{2} } \:rdr$

$\rm :\longmapsto\: {I}^{2} =\pi \displaystyle \int\limits_{0}^{\infty} \: e^{- {r}^{2} } \:2rdr$

On substituting,

$\red{\rm :\longmapsto\: {r}^{2} = t}$

$\red{\rm :\longmapsto\: 2r \: dr = dt}$

So, above integral can be rewritten as

$\rm :\longmapsto\: {I}^{2} =\pi \displaystyle \int\limits_{0}^{\infty} \: e^{- t } \:dt$

$\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg(e^{- t } \bigg)_{ - \infty }^{\infty}$

$\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg(e^{- \infty } - {e}^{0} \bigg)_{ 0 }^{\infty}$

$\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg(0 - 1 \bigg)$

$\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg( - 1 \bigg)$

$\rm :\longmapsto\: {I}^{2} = \pi \:$

$\rm \implies\:\boxed{ \tt{ \: I = \sqrt{\pi} \: }}$

Hence, Proved

Answer 2

Answer:

Hello Peter...

Step-by-step explanation:

Given integral is

\rm :\longmapsto\:\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \: dx:⟼

−∞

∫

∞

e

−x

2

dx

Let assume that,

\rm :\longmapsto\:I = \displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \: dx - - - (1):⟼I=

−∞

∫

∞

e

−x

2

dx−−−(1)

We know, Variable changing property of integrals

\red{\boxed{ \tt{ \: \displaystyle \int\limits_{a}^{b} \: f(x) \: dx \: = \: \displaystyle \int\limits_{a}^{b} \: f(y) \: dy \: }}}

a

∫

b

f(x)dx=

a

∫

b

f(y)dy

So, above integral can also be rewritten as

\rm :\longmapsto\:I = \displaystyle \int\limits_{- \infty}^{\infty} e^{-y^2} \: dy - - - (2):⟼I=

−∞

∫

∞

e

−y

2

dy−−−(2)

Now, multiply equation (1) and (2), we get

\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} dx \times \displaystyle \int\limits_{- \infty}^{\infty} e^{-y^2} \: dy:⟼I

2

=

−∞

∫

∞

e

−x

2

dx×

−∞

∫

∞

e

−y

2

dy

can be rewritten as

\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty}\bigg(\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \times e^{-y^2} \:dx\bigg) \: dy:⟼I

2

=

−∞

∫

∞

(

−∞

∫

∞

e

−x

2

×e

−y

2

dx)dy

\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty}\bigg(\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2 - {y}^{2} } \:dx\bigg) \: dy:⟼I

2

=

−∞

∫

∞

(

−∞

∫

∞

e

−x

2

−y

2

dx)dy

\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{- \infty}^{\infty}\bigg(\displaystyle \int\limits_{- \infty}^{\infty} e^{-(x^2 + {y}^{2})} \:dx\bigg) \: dy:⟼I

2

=

−∞

∫

∞

(

−∞

∫

∞

e

−(x

2

+y

2

)

dx)dy

Now, we can use polar coordinates to evaluate this integral.

Let we substitute

\red{\rm :\longmapsto\: {x}^{2} + {y}^{2} = {r}^{2} }:⟼x

2

+y

2

=r

2

where,

\red{\rm :\longmapsto\:x = r \: cos\theta }:⟼x=rcosθ

and

\red{\rm :\longmapsto\:y = r \: sin\theta }:⟼y=rsinθ

It represents a circle whose center is (0, 0) and radius is r units.

So,

\red{\boxed{ \tt{ \: Limits \:of \: x \: changes \: to \: \theta \: varies \: from \: 0 \: to \: 2\pi \: }}}

Limitsofxchangestoθvariesfrom0to2π

and

\red{\rm :\longmapsto\:dx \: dy \: = \: rd \theta \: dr}:⟼dxdy=rdθdr

So, above integral can be rewritten as

\rm :\longmapsto\: {I}^{2} =\displaystyle \int\limits_{0}^{\infty}\bigg(\displaystyle \int\limits_{0}^{2\pi} d\theta \bigg) \: e^{- {r}^{2} } \:rdr:⟼I

2

=

0

∫

∞

(

0

∫

2π

dθ)e

−r

2

rdr

can be rewritten as

\rm :\longmapsto\: {I}^{2} = \displaystyle \int\limits_{0}^{\infty}\bigg(\theta \bigg)_{0}^{2\pi} \: e^{- {r}^{2} } \:rdr:⟼I

2

=

0

∫

∞

(θ)

0

2π

e

−r

2

rdr

\rm :\longmapsto\: {I}^{2} = \displaystyle \int\limits_{0}^{\infty}\bigg(2\pi - 0 \bigg) \: e^{- {r}^{2} } \:2rdr:⟼I

2

=

0

∫

∞

(2π−0)e

−r

2

2rdr

\rm :\longmapsto\: {I}^{2} = \displaystyle \int\limits_{0}^{\infty}\bigg(2\pi \bigg) \: e^{- {r}^{2} } \:rdr:⟼I

2

=

0

∫

∞

(2π)e

−r

2

rdr

\rm :\longmapsto\: {I}^{2} =\pi \displaystyle \int\limits_{0}^{\infty} \: e^{- {r}^{2} } \:2rdr:⟼I

2

=π

0

∫

∞

e

−r

2

2rdr

On substituting,

\red{\rm :\longmapsto\: {r}^{2} = t}:⟼r

2

=t

\red{\rm :\longmapsto\: 2r \: dr = dt}:⟼2rdr=dt

So, above integral can be rewritten as

\rm :\longmapsto\: {I}^{2} =\pi \displaystyle \int\limits_{0}^{\infty} \: e^{- t } \:dt:⟼I

2

=π

0

∫

∞

e

−t

dt

\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg(e^{- t } \bigg)_{ - \infty }^{\infty}:⟼I

2

=−π(e

−t

)

−∞

∞

\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg(e^{- \infty } - {e}^{0} \bigg)_{ 0 }^{\infty}:⟼I

2

=−π(e

−∞

−e

0

)

0

∞

\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg(0 - 1 \bigg):⟼I

2

=−π(0−1)

\rm :\longmapsto\: {I}^{2} = - \pi \: \bigg( - 1 \bigg):⟼I

2

=−π(−1)

\rm :\longmapsto\: {I}^{2} = \pi \::⟼I

2

=π

\rm \implies\:\boxed{ \tt{ \: I = \sqrt{\pi} \: }}⟹

I=

π

Hence, Proved